Question

[Phạm Hoàng Tiến]

giúp em 3 câu này với ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

Câu 11:

Gọi tọa độ chân đường cao kẻ từ A xuống BC là H(x;y)

=>\(AH\perp\)BC

A(-1;2); B(0;3); C(5;-2)

\(\overrightarrow{BC}=\left(5;-5\right);\overrightarrow{BH}=\left(x;y-3\right)\)

\(\overrightarrow{AH}=\left(x+1;y-2\right)\)

B,H,C thẳng hàng nên ta có: \(\dfrac{x}{5}=\dfrac{y-3}{-5}\)

=>x=-y+3

=>x+y=3(1)

AH\(\perp\)BC

=>\(\overrightarrow{AH}\cdot\overrightarrow{BC}=0\)

=>\(5\left(x+1\right)+\left(-5\right)\cdot\left(y-2\right)=0\)

=>\(\left(x+1\right)-\left(y-2\right)=0\)

=>x+1-y+2=0

=>x-y=-3(2)

Từ (1) và (2) ta có hệ phương trình

\(\left\{{}\begin{matrix}x+y=3\\x-y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x+y=3\end{matrix}\right.\)

=>x=0 và y=3

=>Chọn A

Câu 12: 

B(-1;3); C(3;1); A(x;y)

\(\overrightarrow{AB}=\left(-1-x;3-y\right)\); \(\overrightarrow{AC}=\left(3-x;1-y\right)\)

\(AB=\sqrt{\left(-1-x\right)^2+\left(3-y\right)^2}=\sqrt{\left(y-3\right)^2+\left(x+1\right)^2}\)

\(AC=\sqrt{\left(3-x\right)^2+\left(1-y\right)^2}=\sqrt{\left(x-3\right)^2+\left(y-1\right)^2}\)

ΔABC vuông cân tại A

=>AB\(\perp\)AC và AB=AC

=>\(\left\{{}\begin{matrix}\overrightarrow{AB}\cdot\overrightarrow{AC}=0\\AB=AC\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(-1-x\right)\left(3-x\right)+\left(3-y\right)\left(1-y\right)=0\\\left(y-3\right)^2+\left(x+1\right)^2=\left(x-3\right)^2+\left(y-1\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+1\right)\left(x-3\right)+\left(y-3\right)\left(y-1\right)=0\\y^2-6y+9+x^2+2x+1=x^2-6x+9+y^2-2y+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+1\right)\left(x-3\right)+\left(y-3\right)\left(y-1\right)=0\\-6y+2x=-6x-2y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+1\right)\left(x-3\right)+\left(y-3\right)\left(y-1\right)=0\\8x=4y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x\\x^2-2x-3+y^2-4y+3=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x\\x^2-2x+y^2-4y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x\\x^2-2x+4x^2-4\cdot2x=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x\\5x^2-10x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x\left(x-2\right)=0\\y=2x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(x-2\right)=0\\y=2x\end{matrix}\right.\)

\(x\left(x-2\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Khi x=0 thì \(y=2\cdot0=0\)

Khi x=2 thì \(y=2\cdot2=4\)

=>Chọn B

Câu 13: A(0;4); B(3;4); C(3;0)

\(AB=\sqrt{\left(3-0\right)^2+\left(4-4\right)^2}=3\)

\(AC=\sqrt{\left(3-0\right)^2+\left(0-4\right)^2}=\sqrt{3^2+4^2}=5\)

\(BC=\sqrt{\left(3-3\right)^2+\left(0-4\right)^2}=4\)

Vì \(AB^2+BC^2=AC^2\)

nên ΔABC vuông tại B

=>\(R=\dfrac{AC}{2}=2,5\)

=>Chọn A

Answer 2

[A DUY]

11 a
12 b
13 a